VITEEE 2014 :: Physics :: General questions

• Physics - General questions

In hydrogen discharge tube, it is observed that through a given cross-section $3.31 \times 10^{15}$ electrons are moving from right to left and $3.12 \times 10^{5}$ protons are moving from left to right . The current in the discharge  tube and its direction will be 

Answer:

Explanation:

Here, number of electrons $n_{e}=3.13 \times 10^{15}$

and number of protons $n_{p}=3.12 \times 10^{15}$

 Current $I= n_{e}q_{e} +n_{p}q_{p}$

  $=3.13 \times 10^{15} \times 1.6 \times 10^{-17} +3.12 \times 10^{15}\times 1.6 \times 10^{-19}$

  $=1 \times 10^{-3}=1mA$

Now, due to excess charge on electrons , the direction of the current will be towards right

The resistance of germanium junction diode whose V-I is shown in figure is$ (V_{k}=0.3V)$

Answer:

Explanation:

Resistance , $R=\frac{\triangle V}{\triangle I}=\frac{2.3-0.3}{10 \times 10^{-3}}$

 $R= \frac{2}{10} \times 10^{3}=0.2 \times 10^{3} \Omega=0.2 k \Omega$

Which has more luminous efficiency?

Answer:

Explanation:

Luminous efficiency for the same power supply, 40W fluorescent tube gives more light. Hence, 40W fluorescent tube has a greater

If a magnet is suspened at angle $30^{0}$ to the magnet meridian , the dip of needle makes angle of $45^{0}$ with the horizontal , the real dip is 

Answer:

Explanation:

Hence,  

$tan \delta'=\frac{\tan\delta}{\cos \theta}=\frac{\tan 45^{0}}{\cos 30^{0}}$

$\tan \delta=\frac{1}{\sqrt{3}/2}=\frac{2}{\sqrt{3}}$

or   $\delta=\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$

A neutron is moving with velocity u. It collides head-on and elastically with an atom of mass number A. If the initial kinetic energy of the neutron A. If the initial kinetic energy of the neutron is E, then how much kinetic energy will be retained by the neutron after reflection?

Answer:

Explanation:

Fraction retained by the nucleus 

$\left(\frac{\triangle k}{k}\right)_{retained}=\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)^{2}=\left(\frac{A-1}{A+1}\right)^{2}$

After collision kinetic energy retained by neutron 

$\left(\frac{A-1}{A+1}\right)^{2}E$

• Physics - General questions